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slawek
Posts:
25
Registered:
12/12/08


Re: f(z) + abs(z) = 0, where z is complex
Posted:
Jul 11, 2011 2:10 AM


U?ytkownik "Gib Bogle" napisa? w wiadomo?ci grup dyskusyjnych:ivdv92$384$2@speranza.aioe.org...
>If you can't find an analytical solution to your pair of equations then you >might be out of luck.
1. Analytical solutions are not avaliable.
2. The good approximate values are given "for free".
3. The f(z) is not a simple one. Indeed, the problem was Sum[ a(z,n,m) + b(z,n,m) abs(z)^2 = c(n,m) , n = 1, ..., N ] where m = 1, ..., M .
4. FYI: There is a trouble "what is a solution?" The eq. 1 + abs(x) = 0, where x is a real variable, have got two dinstinct solutions x1 = 1 and x2 = +1. The eq. 1 + abs(z) = 0, where z is complex, have got the circle as the solution. Therefore it is possible to obtain a continuum instead a single number. But the next routine need a number, not a functional dependency.
slawek



